PHYS 232 - UNIVERSITY PHYSICS
II - Temperature and heat
Question Bank - Set 4
Liberty University
Question 1
Question
A steel rod of length 2 m and diameter 5 cm is initially at a uniform temperature
of 100
°
C. It is heated until the temperature is 200
°
C. If the thermal conductivity
of steel is 50 W/m ·K and the specific heat capacity is 500 J/kg ·K, calculate
the heat energy absorbed by the rod during this process.
Solution
Step 1: Calculate the volume of the steel rod. The volume of a cylinder is given
by the formula V=πr2h, where ris the radius and his the height. Given
that the diameter of the rod is 5 cm, the radius ris half of the diameter, so
r=5
2cm = 0.025 m. Therefore, the volume of the rod is:
V=π(0.025)2×2 = 0.001963495 m3
Step 2: Calculate the mass of the steel rod. The density of steel is typically
around 7850 kg/m3. The mass of the rod can be calculated using the formula
m=ρV , where ρis the density. Therefore, the mass of the rod is:
m= 7850 ×0.001963495 = 15.41672275 kg
Step 3: Calculate the change in temperature. The change in temperature of
the rod is 200C−100C= 100C= 100 K.
Step 4: Calculate the heat energy absorbed by the rod. The heat energy
absorbed can be calculated using the formula Q=mc∆T, where mis the mass,
cis the specific heat capacity, and ∆Tis the change in temperature. Therefore,
Q= 15.41672275 ×500 ×100 = 770836.1375 J
So, the heat energy absorbed by the steel rod during this process is 770836.1375 J.
Question 2
Question
A copper rod of length 1 m and initial temperature of 100
°
C is placed in contact
with a large heat reservoir at 0
°
C. The rod is allowed to come to thermal equilib-
rium with the reservoir. If the rod has a thermal conductivity of 385 W/(m
·
K)
and a cross-sectional area of 1 ×10−4m2, determine the rate at which heat is
transferred through the rod during this process.
Solution
Step 1: Calculate the temperature difference across the rod. Given that the
initial temperature of the rod is 100
°
C and it is in contact with a reservoir at
0
°
C, the temperature difference across the rod is:
∆T= 100C−0C= 100C
Step 2: Calculate the rate of heat transfer through the rod using Fourier’s
Law: The rate of heat transfer Qthrough the rod is given by:
Q=k·A·∆T
L
where: - kis the thermal conductivity of copper (385 W/(m
·
K)) - Ais the
cross-sectional area of the rod (1 ×10−4m2)-∆Tis the temperature difference
across the rod (100C) - Lis the length of the rod (1 m)
Plugging in the values:
Q= 385 W/(m
·
K) ×1×10−4m2×100C
1 m
Q= 38.5 W
Therefore, the rate at which heat is transferred through the rod during this
process is 38.5 W.
Question 3
Question
A copper rod of length 2 m has a cross-sectional area of 10 cm2. If one end of
the rod is at 100◦C and the other end is at 50◦C, calculate the rate of heat
conducted along the rod.
Given: Thermal conductivity of copper: 400 W/mK, Density of copper:
8900 kg/m3, Specific heat capacity of copper: 384 J/kgK.
2
Solution
Step 1: Calculate the heat conducted through the rod using Fourier’s law of
heat conduction:
Q=k·A·∆T
L
where: Q= rate of heat transfer (in watts), k= thermal conductivity of the
material (in W/mK), A= cross-sectional area of the material (in m2), ∆T=
change in temperature (in Kelvin), L= length of the material (in meters).
Given: k= 400 W/mK, A= 10 ×10−4m2, ∆T= 100◦C−50◦C = 50 K,
L= 2 m.
Substitute the values into the formula:
Q=400 ·10 ×10−4·50
2= 100 W
Therefore, the rate of heat conducted along the rod is 100 watts.
Question 4
Question
A copper block has a mass of 500 grams and is initially at a temperature of
100
°
C. It is placed in a bucket containing 2 kg of water at 20
°
C. Assuming no
heat is lost to the surroundings, calculate the final equilibrium temperature of
the system when the block and water reach thermal equilibrium. (Specific heat
capacity of copper = 0.386 J/g
°
C, specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the copper block Let Tbe the final
equilibrium temperature of the system. The heat lost by the copper block is
equal to the heat gained by the water:
(mcopper)(ccopper)(T−100) = (mwater)(cwater)(T−20)
Substitute the given values:
(500 g)(0.386 J/g
°
C)(T−100) = (2000 g)(4.18 J/g
°
C)(T−20)
Step 2: Solve for TExpand and simplify the equation:
193T−19300 = 8360T−167200
8360T−193T= 167200 −19300
8167T= 148900
T=148900
8167 ≈18.22C
Therefore, the final equilibrium temperature of the system when the block
and water reach thermal equilibrium is approximately 18.22C.
3
Question 5
Question
A copper bar of length 1.5 m and cross-sectional area 0.02 m2is heated from
20
°
C to 80
°
C. If the thermal conductivity of copper is 400 W/(m·K), calculate
the amount of heat transferred to the copper bar.
Solution
Step 1: Determine the initial and final temperatures of the copper bar in Kelvin.
Initial temperature (in K) = 20C+ 273.15 = 293.15K
Final temperature (in K) = 80C+ 273.15 = 353.15K
Step 2: Calculate the temperature difference (∆T) between the initial and
final temperatures.
∆T= 353.15K−293.15K
= 60K
Step 3: Calculate the heat transferred using the formula:
Q=k·A·∆T·t
where: Q= heat transferred (in Joules)
k= thermal conductivity of copper (400 W/(m·K))
A= cross-sectional area of the bar (0.02 m2)
∆T= temperature difference (60 K)
t= length of time of heat transfer (let’s assume 1 second for simplicity)
Step 4: Substitute the given values into the formula to find the heat trans-
ferred.
Q= 400 W/(m·K) ×0.02 m2×60 K ×1 s
= 480 J
Therefore, the amount of heat transferred to the copper bar is 480 Joules.
Question 6
Question
A metal bar of length 1.5 m and cross-sectional area 0.02 m2has a thermal
conductivity of 70 W/(m*K). One end of the bar is kept at 100◦C while the
other end is kept at 20◦C. Find the rate of heat transfer along the bar.
4
Solution
Step 1: Calculate the temperature difference along the bar. Given that the
metal bar has one end at 100◦C and the other end at 20◦C, the temperature
difference (∆T) is:
∆T= 100 −20 = 80 ◦C
Step 2: Calculate the rate of heat transfer using Fourier’s Law. Fourier’s
Law states that the rate of heat transfer ( ˙
Q) through a material is given by:
˙
Q=−kAdT
dx
where: ˙
Q= rate of heat transfer, k= thermal conductivity of the material, A
= cross-sectional area of the bar, dT
dx = temperature gradient along the bar.
Step 3: Calculate the temperature gradient using the temperature difference
and the length of the bar. The temperature gradient (dT
dx ) is given by:
dT
dx =∆T
L
where Lis the length of the bar. Substitute the values:
dT
dx =80 ◦C
1.5 m =80
1.5
◦C/m = 53.3◦C/m
Step 4: Substitute the values into Fourier’s Law to find the rate of heat
transfer. ˙
Q=−70 ×0.02 ×53.3
˙
Q=−70 ×0.02 ×53.3 = −74.6667 W
Therefore, the rate of heat transfer along the bar is 74.6667 W.
Question 7
Question
A block of metal initially at a temperature of 500 K is heated until its tem-
perature reaches 800 K. If the specific heat capacity of the metal is 0.5 J/g◦C,
calculate the amount of heat absorbed by a 250 g block of the metal during this
process.
Solution
Step 1: Calculate the change in temperature. Given: Initial temperature, T1=
500 K Final temperature, T2= 800 K Change in temperature, ∆T=T2−T1
∆T= 800 K −500 K = 300 K
5
Step 2: Calculate the amount of heat absorbed using the formula:
Q=mc∆T
where: Q= heat absorbed m= mass of the block = 250 g = 0.25 kg c=
specific heat capacity of the metal = 0.5 J/g◦C= 500 J/kg◦C
Substitute the values into the formula:
Q= 0.25 kg ×500 J/kg◦C×300 K
Step 3: Calculate the value of Q.
Q= 0.25 ×500 ×300 = 37,500 J
Therefore, the amount of heat absorbed by the block of metal is 37,500 J.
Question 8
Question
A block of copper with a mass of 0.5 kg and a specific heat capacity of 0.385
J/(g·
°
C) is initially at a temperature of 100
°
C. It is placed in a container with
2 kg of water at an initial temperature of 20
°
C. If the final temperature of the
system is 25
°
C, what is the specific heat capacity of water?
Solution
Step 1: Find the heat absorbed by the copper block using the formula:
Q=mc∆T
where: m= 0.5 kg (mass of copper), c= 0.385 J/(g ·
°
C) (specific heat capacity
of copper), ∆T= 25 −100 = −75
°
C (change in temperature).
Plugging in the values, we get:
Qcopper = (0.5 kg)(0.385 J/(g ·
°
C))(−75
°
C)
Qcopper =−14.375 J
Step 2: Find the heat released by the water when it cools down using the
same formula:
Q=mc∆T
where: m= 2 kg (mass of water), cis the specific heat capacity of water (to be
determined), ∆T= 25 −20 = 5
°
C (change in temperature).
Plugging in the values, we have:
−Qwater = (2 kg)(c)(5
°
C)
6
−Qwater = 10cJ
Step 3: Since the total heat gained by the water is equal to the total heat
lost by the copper block (by conservation of energy), we have:
Qcopper =Qwater
−14.375 J = 10cJ
c=−1.4375 J/(g ·
°
C)
Therefore, the specific heat capacity of water is 1.4375 J/(g·
°
C).
Question 9
Question
A block of copper initially at 100
°
C is dropped into a vessel containing 1 kg of
water at 20
°
C. The specific heat capacity of copper is 0.385 J/g
°
C and that of
water is 4.18 J/g
°
C. The block of copper comes to thermal equilibrium with the
water. Calculate the final temperature of the system.
Solution
Step 1: Find the heat lost by the copper block.
The heat lost by the copper block can be calculated using the formula:
Q=mc∆T, where: - mis the mass of the copper block, - cis the specific heat
capacity of copper, and - ∆Tis the change in temperature.
Given: - Initial temperature of copper block (Tcopper) = 100
°
C, - Final
temperature of system = Tf, - Specific heat capacity of copper (ccopper ) =
0.385 J/g
°
C, - Mass of copper block = mcopper, - Change in temperature (∆Tcopper)
=Tcopper −Tf.
The heat lost by the copper block is equal to the heat gained by the water,
so we have: mccopper ∆Tcopper =mcwater ∆Twater
Step 2: Find the heat gained by the water.
The heat gained by the water can be calculated using the same formula:
Q=mc∆T, where: - mis the mass of the water, - cis the specific heat capacity
of water, and - ∆Tis the change in temperature.
Given: - Initial temperature of water (Twater) = 20
°
C, - Mass of water = 1
kg = 1000 g, - Specific heat capacity of water (cwater) = 4.18 J/g
°
C, - Change
in temperature (∆Twater ) = Tf−Twater .
Step 3: Set up and solve the equations to find the final temperature.
From Step 1: mcopperccopper (Tcopper −Tf) = mwater cwater (Tf−Twater )
Substitute the given values and solve for Tfto find the final temperature of
the system.
7
Question 10
Question
A copper rod of length 2 m and diameter 1 cm conducts heat at a rate of 200 W
when its ends are at temperatures of 100◦C and 0◦C. The thermal conductivity
of copper is 400 W/m·K. How long does it take for the rod to reach a uniform
temperature of 50◦C?
Solution
Step 1: Find the cross-sectional area of the rod. The diameter of the rod is
1 cm, so the radius is 0.5 cm = 0.005 m. The cross-sectional area A of the
rod is given by the formula A=πr2, where r is the radius. A=π(0.005)2=
π×0.000025 = 7.85398 ×10−5m2.
Step 2: Find the rate of heat conduction per unit area. The rate of heat
conduction Q through a material is given by Fourier’s law: Q=kA∆T /L,
where k is the thermal conductivity, ∆Tis the temperature gradient, and L is
the length of the material. Given that Q = 200 W, k = 400 W/m·K, ∆T=
100 −0 = 100 ◦C = 100 K, and L = 2 m, we can rearrange the formula to find
Q
A=k∆T
L. Substitute the given values to find 200
7.85398×10−5= 400 ×100
2. This
simplifies to 200
7.85398×10−5= 200 ×104= 2 ×106W/m2.
Step 3: Find the time taken to reach a uniform temperature of 50◦C. The
rate of temperature change dT
dt at a point along the rod is given by Fourier’s law
in differential form: Q=−kA dT
dx , where x is the distance, in this case, along
the rod. At the midpoint of the rod, x = L/2 = 1 m. Solving for dT
dt , we get
dT
dt =−Q
kA =−2×106
400×7.85398×10−5=−50661430.7 K/s. To find the time taken
for the rod to reach a uniform temperature of 50◦C = 50 K, divide ∆Tby dT
dt .
The time taken is 50
50661430.7≈9.88 ×10−7s.
Therefore, it takes approximately 9.88 ×10−7seconds for the rod to reach
a uniform temperature of 50◦C.
Question 11
Question
A 150 g cube of ice at -10
°
C is placed in a 300 g aluminum calorimeter cup
of heat capacity 0.40 J/g
°
C containing 500 g of water at 20
°
C. Assume that
no heat is lost to the surroundings. What will be the final temperature of the
system?
Given: Specific heat capacity of ice: 2.09 J/g
°
C Specific heat capacity of
water: 4.18 J/g
°
C Heat of fusion of ice: 334 J/g
8
Solution
Step 1: Calculate the heat lost by the ice to reach 0
°
C. The heat lost by the ice
is given by the formula:
Qice =mice ·cice ·(0 −(−10))
Substitute the values:
Qice = 150 g ·2.09 J/g
°
C·10
°
C
Calculating:
Qice = 3135 J
Step 2: Calculate the heat gained by the ice at 0
°
C to melt into water. The
heat gained by the ice to melt is given by the formula:
Qmelt =mice ·Heat of fusion of ice
Substitute the given values:
Qmelt = 150 g ·334 J/g
Calculating:
Qmelt = 50100 J
Step 3: Calculate the final temperature after the ice has melted and reached
thermal equilibrium with the water. Let the final temperature be Tf.
Using the conservation of energy equation:
Qice +Qmelt = (mice ·cice +mcalorimeter ·ccalorimeter +mwater ·cwater)·(Tf−20)
Substitute the known values:
3135 + 50100 = (150 ·2.09 + 300 ·0.40 + 500 ·4.18) ·(Tf−20)
Solve for Tf:
53235 = (313.5 + 120 + 2090) ·(Tf−20)
53235 = 3523.5·(Tf−20)
Tf−20 = 53235
3523.5
Tf−20 ≈15
°
C
Therefore, the final temperature of the system is approximately 15
°
C.
9
Question 12
Question
A copper block with a mass of 0.5 kg is heated from an initial temperature of
25
°
C to a final temperature of 75
°
C. If the specific heat capacity of copper is
390 J/kg◦C, calculate the amount of heat energy absorbed by the block during
this process.
Solution
Step 1: Determine the change in temperature. Given: Initial temperature (Ti)
= 25
°
C, Final temperature (Tf) = 75
°
C.
The change in temperature (∆T) is calculated as:
∆T=Tf−Ti= 75◦C−25◦C = 50◦C
Step 2: Calculate the amount of heat energy absorbed. The amount of heat
energy absorbed (Q) is given by the formula:
Q=mc∆T
where: m= mass of the block = 0.5 kg, c= specific heat capacity of copper =
390 J/kg◦C.
Substitute the values into the formula:
Q= (0.5 kg)(390 J/kg◦C)(50◦C)
Q= 9750 J
Therefore, the amount of heat energy absorbed by the block during this
process is 9750 Joules.
Question 13
Question
A copper sphere with a radius of 5 cm is heated from an initial temperature
of 20
°
C to a final temperature of 80
°
C. If the specific heat capacity of copper
is 0.385 J/g
°
C and its density is 8.96 g/cm3, calculate the heat supplied to the
sphere in this process. Assume the sphere is perfectly insulated and no heat is
lost to the surroundings.
Solution
Step 1: Calculate the mass of the copper sphere using its density and volume.
Volume of sphere = 4
3πr3
10
Volume of sphere = 4
3π(5 cm)3
Volume of sphere = 4
3π×125 cm3
Volume of sphere = 500
3πcm3≈523.6 cm3
Mass = Volume ×Density
Mass = 523.6 cm3×8.96 g/cm3
Mass ≈4684.3 g = 4.6843 kg
Step 2: Calculate the change in temperature of the copper sphere.
∆T= 80C−20C= 60C
Step 3: Determine the heat supplied to the copper sphere using the formula:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
Q= 4.6843 ×0.385 ×60
Q≈108.01 kJ
Therefore, the heat supplied to the copper sphere in this process is approx-
imately 108.01 kJ.
Question 14
Question
A copper cylinder of mass 2 kg contains 0.5 kg of water at a temperature of
100◦C. The copper cylinder is then heated to a temperature of 200◦C. If the
specific heat capacity of copper is 390 J/kg·K and the specific heat capacity of
water is 4186 J/kg·K, determine the final temperature of the water and copper
cylinder system.
Solution
Step 1: Calculate the heat absorbed or released by the water using the formula:
Q=mc∆T
where Q is the heat energy, m is the mass, c is the specific heat capacity, and
∆Tis the change in temperature.
Given: - mwater = 0.5 kg - cwater = 4186 J/kg·K - Initial temperature of
water, Tin = 100◦C - Final temperature of water, Tfin
11
The change in temperature (∆T) for water is:
∆Twater =Tfin −Tin
The heat absorbed by water is:
Qwater =mwatercwater∆Twater
Step 2: Calculate the heat absorbed or released by the copper cylinder using
the same formula:
Q=mc∆T
Given: - mcopper = 2 kg - ccopper = 390 J/kg·K - Initial temperature of
copper, Tin = 100◦C - Final temperature of copper, Tfin = 200◦C
The change in temperature (∆T) for copper is:
∆Tcopper =Tfin −Tin
The heat absorbed by copper is:
Qcopper =mcopperccopper∆Tcopper
Step 3: Since the total heat energy absorbed by the water must equal the
total heat energy absorbed by the copper cylinder, we can set up an equation:
Qwater =Qcopper
Step 4: Substitute the expressions for Qwater and Qcopper and solve for Tfin.
Question 15
Question
A metal bar of length 2 m and cross-sectional area 0.02 m2is initially at a
uniform temperature of 200◦C. The bar is then placed in an environment at
20◦C, causing it to cool. If the thermal conductivity of the metal is 50 W/(m·K)
and the heat transfer coefficient between the bar and surrounding air is 25
W/(m2·K), determine the time it takes for the bar to cool to 100◦C. Assume
that the surroundings are at a constant temperature of 20◦C.
Solution
Step 1: Calculate the heat transfer rate from the bar to the surroundings using
Newton’s Law of Cooling: The heat transfer rate (Q) is given by:
Q=hA(T−Tsurroundings)
where h= heat transfer coefficient = 25 W/(m2·K), A= cross-sectional area =
0.02 m2,T= temperature of the bar at any time t,Tsurroundings = temperature
of the surroundings = 20◦C.
12
Step 2: Calculate the thermal resistance of the bar: The thermal resistance
(R) is given by:
R=L
kA
where L= length of the bar = 2 m, k= thermal conductivity = 50 W/(m·K),
A= cross-sectional area = 0.02 m2.
Step 3: Use the concept of thermal resistance to express the rate of tem-
perature change of the metal bar: The rate of temperature change is given
by: dT
dt =−Q
ρcV
where ρ= density of the material, c= specific heat capacity of the material, V
= volume of the bar.
Step 4: Substitute the expressions for Qand dT
dt into the rate of temperature
change equation and integrate to find the time tit takes for the bar to cool to
100◦C:
ZTf
Ti
dT
T−Tsurroundings
=−2
50 ·0.02 ·ρcV Zt
0
dt
where Ti= initial temperature of the bar = 200◦C, Tf= final temperature of
the bar = 100◦C.
Question 16
Question
A 0.5 kg aluminum pot containing 1 kg of water at 20
°
C is placed on a stove.
The stove supplies heat at a rate of 500 W. Assuming no heat is lost to the
surroundings, calculate how long it will take for the water to boil (100
°
C). The
specific heat capacity of aluminum is 900 J/kg◦C and the specific heat capacity
of water is 4186 J/kg◦C.
Solution
Step 1: Calculate the heat required to raise the temperature of water from 20
°
C
to 100
°
C. The heat Qrequired to raise the temperature of a material can be
calculated using the formula:
Q=mc∆T
where: - mis the mass of the material, - cis the specific heat capacity of the
material, - ∆Tis the change in temperature.
For the water:
Qwater = (1 kg)(4186 J/kg◦C)(80 ◦C)
Qwater = 334880 J
13
Step 2: Calculate the heat required to raise the temperature of the aluminum
pot from room temperature to the boiling point of water. For the aluminum
pot:
Qaluminum = (0.5 kg)(900 J/kg◦C)(80 ◦C)
Qaluminum = 36000 J
Step 3: Calculate the total heat provided by the stove to boil the water. The
total heat provided is the sum of the heat required for both the water and the
aluminum pot.
Qtotal =Qwater +Qaluminum
Qtotal = 334880 J + 36000 J = 370880 J
Step 4: Calculate the time taken for the water to boil. The rate of heat
supplied by the stove is given as 500 W (J/s). We can use the formula:
Power = Energy
Time
500 W = 370880 J
t
Solving for time t:
t=370880 J
500 W
t= 741.76 s
Therefore, it will take approximately 741.76 seconds for the water to boil.
Question 17
Question
A 2 kg block of ice at −20◦Cis placed in an insulated container holding 10 kg
of water at 25◦C. Assuming no heat is exchanged with the surroundings, what
will be the final temperature of the system? (Specific heat capacity of water =
4186 J/(kg ·K) and specific heat capacity of ice = 2100 J/(kg ·K)).
Solution
Step 1: First, we need to calculate the heat lost by the ice and the heat gained
by the water until they reach thermal equilibrium. The heat lost by the ice can
be calculated as:
Qice =mice ·cice ·(Tf−Tice)
where: - mice = 2 kg (mass of ice), - cice = 2100
Question
A copper rod with a length of 1.5 m and a diameter of 2 cm is initially at a
temperature of 200
°
C. It is placed in a room where the temperature is 25
°
C. If
14
the thermal conductivity of copper is 390 W/mK, the specific heat capacity is
390 J/kgK, and the density is 8900 kg/m3, calculate the time it takes for the
rod to reach thermal equilibrium with the room. Assume one end of the rod is
perfectly insulated.
Solution
Step 1: Calculate the surface area of the rod. The surface area of a rod can be
calculated using the formula A=π×r×h, where ris the radius and his the
length of the rod. Given that the diameter of the rod is 2 cm, the radius ris 1
cm or 0.01 m. So, A=π×0.01 ×1.5 = 0.0471 m2.
Step 2: Calculate the volume of the rod. The volume of the rod can be
calculated using the formula V=π×r2×h. Given that the diameter of the
rod is 2 cm, the radius ris 1 cm or 0.01 m. So, V=π×(0.01)2×1.5 = 4.712
x 10−4m3.
Step 3: Calculate the mass of the rod. The mass of the rod can be calculated
using the formula m= density ×volume. Given that the density of copper is
8900 kg/m3, the mass m= 8900 ×4.712 x 10−4= 4.2 kg.
Step 4: Calculate the total energy required to heat the rod from 25
°
C to
200
°
C. The total energy Qrequired can be calculated using the formula Q=
mc∆T, where mis the mass, cis the specific heat capacity, and ∆Tis the
temperature difference. The temperature difference ∆T= 200 −25 = 175
°
C.
So, Q= 4.2×390 ×175 = 296550 J.
Step 5: Calculate the rate of heat transfer through the rod. The rate of heat
transfer Pcan be calculated using Fourier’s law: P=kA∆T/L, where kis the
thermal conductivity, Ais the surface area, ∆Tis the temperature difference,
and Lis the length of the rod. One end of the rod is insulated, so only one side is
in contact with the room. Therefore, L= 1.5 m. So, P= 390×0.0471×175/1.5
= 21.64 W.
Step 6: Calculate the time taken for the rod to reach thermal equilibrium.
The time ttaken can be calculated using the formula Q=P t, where Qis the
total energy required, and Pis the rate of heat transfer. So, t= 296550/21.64
= 13706 seconds or 3.81 hours. Therefore, it takes approximately 3.81 hours for
the rod to reach thermal equilibrium with the room.
Question 19
Question
A piece of aluminum initially at 200
°
C is placed in a lake with a temperature of
10
°
C. If the specific heat capacity of aluminum is 0.897 J/g
°
C, and the mass of
the aluminum is 500 g, calculate the amount of heat that must be transferred
to the aluminum to bring it to thermal equilibrium with the lake.
15
Solution
Step 1: Calculate the change in temperature of the aluminum. Let Tfbe the
final temperature of the aluminum. The change in temperature ∆Tis given by:
∆T=Tf−Ti
where Ti= 200Cis the initial temperature of the aluminum. Therefore, ∆T=
Tf−200.
Step 2: Calculate the heat absorbed or lost by the aluminum using the
formula:
q=mc∆T
where qis the heat, mis the mass of the aluminum, cis the specific heat capacity
of aluminum, and ∆Tis the change in temperature.
Step 3: Set up the equation using the given values:
q= (500 g)(0.897 J/g
°
C)(Tf−200)
Step 4: At thermal equilibrium, the total heat loss by the aluminum equals
the total heat gained by the lake. This means:
q=mc∆T=−qlake
where qlake is the heat lost by the lake (negative since the lake gains heat).
Step 5: Since the lake is at 10
°
C, the heat lost by the lake is given by:
−qlake = 500 g ×(specific heat capacity of water) ×(Tf−10)
Step 6: Equate qand −qlake:
500 ×0.897 ×(Tf−200) = −500 ×(specific heat capacity of water) ×(Tf−10)
Step 7: Solve for Tfto find the final temperature of the aluminum when it
reaches thermal equilibrium with the lake. After finding Tf, substitute it back
into the equation for qto calculate the amount of heat transferred.
Question 20
Question
A metal bar of length 2 m and cross-sectional area 0.01 m2is heated at one end
and the other end is kept at a constant temperature. The thermal conductivity
of the metal is 200 W/(m K). If the temperature gradient along the length of
the bar is 100 K/m, determine the rate at which energy is conducted along the
bar.
16
Solution
Let’s denote the rate at which energy is conducted along the bar by Q. We can
use Fourier’s law of heat conduction which states:
Q=−kA∆T
∆x
where: Q= rate of heat conduction, k= thermal conductivity of the material, A
= cross-sectional area, ∆T= change in temperature, ∆x= change in position.
Step 1: Find the change in position ∆x.Given that the length of the
bar is 2 m and the temperature gradient is 100 K/m, we can find the change in
position as:
∆x= 2 m
Step 2: Calculate the rate of heat conduction. Plugging in the values
into Fourier’s law of heat conduction, we get:
Q=−200 ×0.01 ×100 = −200 W
Therefore, the rate at which energy is conducted along the bar is 200 W.
Question 21
Question
A copper block of mass 0.5 kg at an initial temperature of 200◦C is placed in
a calorimeter containing 0.2 kg of water at 20◦C. If the final temperature of
the system is 50◦C, calculate the specific heat capacity of copper. Assume the
specific heat capacity of water is 4200 J/kg·K.
Solution
Step 1: Calculate the heat gained by the water. The heat gained by the water
can be calculated using the formula:
Qwater =m·c·∆T
where: - mis the mass of the water, which is 0.2 kg, - cis the specific heat
capacity of water, which is 4200 J/kg·K, - ∆Tis the change in temperature of
the water, which is 50◦C - 20◦C = 30◦C.
Substitute the given values into the formula:
Qwater = 0.2 kg ×4200 J/kg·K×30 ◦C
Step 2: Calculate the heat lost by the copper block. The heat lost by the
copper block can also be calculated using the formula:
Qcopper =m·c·∆T
17
where: - mis the mass of the copper block, which is 0.5 kg, - cis the specific
heat capacity of copper (to be calculated), - ∆Tis the change in temperature of
the copper block, which is 50◦C - 200◦C = -150◦C (negative because the block
is losing heat).
Step 3: The total heat lost by the copper block is equal to the total heat
gained by the water.
Qcopper =Qwater
mc·cc·∆Tc=mw·cw·∆Tw
Solve for ccto find the specific heat capacity of copper:
cc=mw·cw·∆Tw
mc·∆Tc
Substitute the known values into the formula and solve for cc.
Question 22
Question
A 10 kg block of copper initially at a temperature of 100◦C is dropped into a
container of 20 kg of water at an initial temperature of 20◦C. Assuming no heat
loss to the surroundings, calculate the final equilibrium temperature of the block
and water system. (Specific heat capacity of copper = 390 J/kg◦C, specific heat
capacity of water = 4186 J/kg◦C)
Solution
Step 1: Calculate the heat lost by the copper block as it cools down to the final
temperature of the system. The heat lost by an object can be calculated using
the formula:
Q=mc∆T
where: Q= heat lost, m= mass of the object, c= specific heat capacity of the
object, ∆T= change in temperature.
For the copper block: m= 10 kg, c= 390 J/kg◦C, Initial temperature =
100◦C, Final temperature = T(since the final equilibrium temperature is not
known yet).
We can then calculate the heat lost by the copper block:
Qcopper = 10 ×390 ×(100 −T)
Step 2: Calculate the heat gained by the water as it heats up to the final
temperature of the system. The heat gained by an object can be calculated
using the same formula as in Step 1.
For the water: m= 20 kg, c= 4186 J/kg◦C, Initial temperature = 20◦C,
Final temperature = T.
18
We can calculate the heat gained by the water:
Qwater = 20 ×4186 ×(T−20)
Step 3: Since the system reaches thermal equilibrium, the heat lost by the
copper block is equal to the heat gained by the water.
Qcopper =Qwater
10 ×390 ×(100 −T) = 20 ×4186 ×(T−20)
Now, we can solve this equation to find the final equilibrium temperature T
of the system.
Question 23
Question
A metal bar of length 2.0 m and cross-sectional area 5.0 cm2is initially at a
temperature of 100◦C. If 5.0 kJ of heat are added to the bar, what is the final
temperature of the bar? Assume the specific heat capacity of the metal is 0.450
kJ/kg◦C and the density is 8.0 g/cm3.
Solution
Step 1: Calculate the mass of the metal bar using its length, cross-sectional
area, and density. The volume of the bar is given by:
V= length ×cross-sectional area = 2.0×5.0×10−4= 0.001 m3
The mass can be calculated from the density:
mass = density ×volume = 8.0×103×0.001 = 8.0 kg
Step 2: Calculate the change in temperature of the metal bar. The heat
added to the bar is given as 5.0 kJ, and the specific heat capacity is 0.450
kJ/kg◦C. So, the change in temperature can be calculated using the formula:
Q=mc∆T
where: Q= heat added = 5000 J m= mass = 8.0 kg c= specific heat capacity
= 450 J/kg◦C
Solving for ∆T:
5000 = 8.0×450 ×∆T
∆T=5000
8.0×450 = 1.389 ◦C
Step 3: Calculate the final temperature of the metal bar. The final temper-
ature can be found by adding the change in temperature to the initial temper-
ature:
Final temperature = 100 + 1.389 = 101.39 ◦C
Therefore, the final temperature of the metal bar after adding 5.0 kJ of heat
is 101.39◦C.
19
Question 24
Question
A copper rod of length 1.5 m and diameter 1 cm is initially at a temperature of
25
°
C. If one end of the rod is immersed in steam at 100
°
C and the other end is
surrounded by ice at 0
°
C, what is the rate at which energy is transferred along
the rod? Given that the thermal conductivity of copper is 390 W/mK.
Solution
Step 1: First, we calculate the cross-sectional area of the rod. The radius of
the rod, r=1 cm
2= 0.005 m. Therefore, the cross-sectional area of the rod,
A=πr2=π(0.005)2m2.
Step 2: Next, we calculate the temperature difference across the rod. The
temperature difference across the rod, ∆T= (100 −25) −(25 −0) = 75
°
C.
Step 3: Now, we can calculate the rate of energy transferred along the rod
using Fourier’s Law of Heat Conduction:
dQ
dt =−kAdT
dx
where, dQ/dt = rate of heat transfer, k= thermal conductivity of copper, A=
cross-sectional area of the rod, dT/dx = temperature gradient along the rod.
Step 4: The temperature gradient along the rod, dT /dx =∆T
L=75
1.5
°
C/m.
Step 5: Substituting the given values into Fourier’s Law, we get:
dQ
dt =−390 ×π×(0.005)2×75
1.5
Step 6: Simplifying,
dQ
dt =−390 ×π×0.000025 ×50
Step 7: Finally, calculating the rate of energy transferred gives:
dQ
dt =−π×0.4875
Therefore, the rate at which energy is transferred along the rod is approxi-
mately −1.53 W/m.
Question 25
Question
A metal rod of length 1 m and diameter 2 cm is initially at a temperature of
100◦C. It is then immersed in a large tank of water at 20◦C. If the rod loses heat
at a rate of 50 W, how long will it take for the rod to cool down to 50◦C? Assume
that the thermal conductivity of the metal is 100 W/mK and the specific heat
capacity of the metal is 500 J/kgK.
20
Solution
Step 1: Calculate the surface area of the rod. The surface area of a cylinder can
be calculated using the formula:
A= 2πrh +πr2
where ris the radius of the cylinder and his its height. In this case, h= 1 m
and r= 1 cm = 0.01 m.
So, the surface area of the rod is:
A= 2π(0.01)(1) + π(0.01)2= 0.04 m2
Step 2: Determine the rate of temperature change. The rate at which the
rod loses heat is given as 50 W. We can calculate this rate using the formula:
Q=kA∆T
∆x
where Qis the heat flow rate, kis the thermal conductivity of the metal, Ais
the surface area of the rod, ∆Tis the temperature difference, and ∆xis the
length of the rod.
Substitute the given values into the equation:
50 = (100)(0.04)(100 −20)
1
Step 3: Calculate the time taken to cool to 50◦C. First, we need to calculate
the mass of the rod using the density formula:
Density = m
Volume
m= Density ×Volume
The volume of the rod is given by:
V=πr2h=π(0.01)2(1) = 0.0001πm3
So, the mass of the rod is:
m= 8000 ×0.0001π= 0.25πkg
The specific heat formula is:
Q=mc∆T
50 ×t= 0.25π×500 ×(100 −50)
50t= 6250π
t≈39.8 s
Therefore, it will take approximately 39.8 seconds for the rod to cool down
to 50◦C.
21
Question 26
Question
A copper cubical block with sides of length 0.1 m is initially at a temperature
of 100
°
C. It is dropped into a large container of cold water at 10
°
C. If the
specific heat capacity of copper is 390 J/kg ·
°
C and the density of copper is
8,960 kg/m3, calculate the total heat transfer when the block reaches thermal
equilibrium with the water. Assume no heat is lost to the surroundings.
Solution
Step 1: Calculate the mass of the copper block. Given that the density of
copper is 8,960 kg/m3and the volume of the copper block is (0.1 m)3, we can
calculate the mass using the formula m=ρV .m= 8,960 kg/m3×(0.1 m)3
m= 8,960 kg/m3×0.001 m3m= 8.96 kg
Step 2: Calculate the heat lost by the copper block. The heat lost by the
copper block can be calculated using the formula Q=mc∆T, where mis the
mass, cis the specific heat capacity, and ∆Tis the change in temperature.
∆T= 100C−10C= 90C Q = 8.96 kg ×390 J/kg ·
°
C×90
°
CQ= 316,872 J
Step 3: Calculate the heat gained by the water. The heat gained by the
water can be calculated using the same formula Q=mc∆T, where the mass
of the water and the specific heat capacity of water are used. Let’s assume the
mass of the water is equal to the mass of the copper block. Q= 8.96 kg ×
4180 J/kg ·
°
C×90
°
CQ= 337,843.2 J
Step 4: Total heat transfer. Since the heat lost by the copper block is equal
to the heat gained by the water, the total heat transfer is the sum of these two
amounts. Total heat transfer = 316,872 J + 337,843.2 J = 654,715.2 J
Therefore, the total heat transfer when the copper block reaches thermal
equilibrium with the water is 654,715.2 J.
Question 27
Question
A copper vessel of mass 500 g contains 2 kg of water at 30
°
C. How much ice at
-10
°
C must be added to it so that the final temperature of the mixture becomes
10
°
C?
Solution
Step 1: Calculate the heat lost by the water to cool down from 30
°
C to 10
°
C.
Let mwbe the mass of the water, Tibe the initial temperature of the water,
Tfbe the final temperature of the water, and cwbe the specific heat capacity
of water. The specific heat capacity of water is 4200 J/kg
°
C. The formula for
22
calculating heat transfer is given by:
Q=mc∆T
where: - Qis the heat transfer (in Joules), - mis the mass (in kg), - cis the
specific heat capacity (in J/kg
°
C), and - ∆Tis the change in temperature (in
°
C).
The heat lost by the water is equal to the heat gained by the ice. Therefore,
we can write:
mwcw(Tf−Ti) = miLf+mici(Tf−Ti)
where: - miis the mass of the ice added, - Lfis the latent heat of fusion of ice
(336,000 J/kg), and - ciis the specific heat capacity of ice (2100 J/kg
°
C).
Step 2: Substituting the known values into the equation, we get:
mw·4200 ·(10 −30) = mi·336000 + mi·2100 ·(10 −(−10))
Step 3: Simplify the equation to solve for mi:
2·4200 ·20 = mi·336000 + mi·2100 ·20
Step 4: Solve for mito find the mass of ice needed to be added to the
mixture.
Question 28
Question
A steel rod of length 2 m and diameter 1 cm is heated from 20
°
C to 100
°
C. If the
coefficient of linear expansion of steel is 12 ×10−6per degree Celsius, calculate
the change in length of the rod.
Solution
Step 1: First, we calculate the initial volume of the steel rod. The initial volume
Vican be calculated using the formula for the volume of a cylinder:
Vi=πd
22
·L,
where dis the diameter of the rod and Lis the length of the rod. Substituting
d= 1 cm, L= 2 m, we get:
Vi=π1
22
·2 = π
2m3.
Step 2: Next, we calculate the final volume of the steel rod. We know that
volume is conserved, so the final volume Vfis equal to the initial volume Vi.
23
The final volume Vfcan be calculated using the formula for the volume of a
cylinder with the new length Lfand the same diameter:
Vf=πd
22
·Lf.
Since Vi=Vf, we can write:
π
2=π1
22
·Lf,
Lf= 4 m.
Step 3: Finally, we calculate the change in length of the rod. The change in
length ∆Lis given by:
∆L=Lf−Li,
∆L= 4 m −2 m,
∆L= 2 m.
Therefore, the change in length of the steel rod when heated from 20
°
C to
100
°
C is 2 meters.
Question 29
Question
A certain object is initially at a temperature of 300 K. If heat is added to
the object at a rate of 50 J/s for 10 minutes, what is the final temperature of
the object? Assume no heat is lost to the surroundings and the specific heat
capacity of the object is 500 J/kg-K.
Solution
Step 1: First, we need to convert the time from minutes to seconds. There are
60 seconds in a minute, so 10 minutes is equal to 600 seconds.
Step 2: Next, we calculate the total heat added to the object using the
formula for heat: Q= rate of heat transfer ×time. Substituting in the known
values, we get:
Q= 50 J/s ×600 s = 30,000 J
Step 3: The change in temperature of the object can be calculated using the
formula ∆T=Q
mc , where mis the mass of the object and cis the specific heat
capacity. Since the mass is not given and does not affect the final temperature,
we can proceed with the calculation:
∆T=30,000 J
m×500 J/kg-K
24
Step 4: The final temperature can be found by adding the change in tem-
perature to the initial temperature. Thus, the final temperature is:
Tfinal = 300 K + ∆T
Step 5: Substituting the value of ∆Tcalculated in step 3 into the equation
from step 4, we get:
Tfinal = 300 K + 30,000 J
m×500 J/kg-K
Question 30
Question
A copper rod of length 1 m and cross-sectional area 0.01 m2is initially at
100◦C. The rod is placed in an ice-water mixture at 0◦C and allowed to reach
thermal equilibrium with the surroundings. Given that copper has a specific
heat capacity of 390 J/(kg·K) and a density of 8930 kg/m3, calculate the amount
of heat transferred from the rod to the surroundings. Assume no heat is lost to
the surroundings other than through the ends of the rod, and that the specific
heat capacity and density of copper remain constant over the temperature range.
Solution
Step 1: Calculate the mass of the copper rod. The mass of the rod can be
calculated using the formula:
mass = density ×volume
Given that the density of copper is 8930 kg/m3and the volume of the rod is
length ×cross-sectional area, we have:
mass = 8930 kg/m3×1 m ×0.01 m2= 89.3 kg
Step 2: Calculate the initial thermal energy of the rod. The initial thermal
energy Qiof the rod can be calculated using the formula:
Qi= mass ×specific heat capacity ×∆T
where ∆Tis the change in temperature. The rod is initially at 100◦C, so
∆T= 100 −0 = 100 K. Substitute the values to get:
Qi= 89.3 kg ×390 J/(kg ·K) ×100 K = 349470 J
Step 3: Calculate the final thermal energy of the rod. When the rod reaches
thermal equilibrium at 0◦C, the final thermal energy Qfof the rod is:
Qf= mass ×specific heat capacity ×∆T
25
where ∆T= 0 −100 = −100 K. Substitute the values to get:
Qf= 89.3 kg ×390 J/(kg ·K) ×(−100) K = −349470 J
Step 4: Calculate the amount of heat transferred. The amount of heat
transferred from the rod to the surroundings can be calculated as the decrease
in thermal energy:
Q=Qf−Qi=−349470 J −349470 J = −698940 J
Therefore, the amount of heat transferred from the rod to the surroundings is
698940 J.
Question 31
Question
A copper cylinder of mass 0.2 kg is initially at 100
°
C. It is placed in a large heat
sink at 0
°
C. If the specific heat capacity of copper is 390 J/kg◦C, calculate the
total heat transfer when the cylinder reaches thermal equilibrium with the heat
sink.
Solution
Step 1: Calculate the change in temperature of the copper cylinder. The heat
transferred by the copper cylinder can be given by the formula:
Q=mc∆T,
where: Q= heat transferred, m= mass of copper cylinder = 0.2 kg, c= specific
heat capacity of copper = 390 J/kg◦C, ∆T= change in temperature.
Initially, the temperature of the copper cylinder is 100
°
C, and the final tem-
perature is 0
°
C. Therefore, the change in temperature is:
∆T= 0 −100 = −100 ◦C.
Step 2: Calculate the heat transferred. Substitute the values into the formula
for heat transferred:
Q= (0.2)(390)(−100).
Calculating this gives:
Q=−7800 J.
Step 3: Interpretation. The negative sign indicates that heat is being trans-
ferred out of the copper cylinder to the heat sink since the final temperature is
lower than the initial temperature. Therefore, the total heat transfer when the
cylinder reaches thermal equilibrium with the heat sink is 7800 J.
26
Question 32
Question
An aluminum rod of length 1 m and cross-sectional area 0.01 m2is initially at
a temperature of 100◦C. The rod is heated until its temperature reaches 200◦C.
The specific heat capacity of aluminum is 900 J/(kg·K) and its density is 2700
kg/m3. How much heat is added to the rod during this process?
Solution
Step 1: Calculate the mass of the aluminum rod. Given the length of the rod
(l) is 1 m, the cross-sectional area (A) is 0.01 m2, and the density of aluminum
(ρ) is 2700 kg/m3, we can calculate the mass (m) of the rod using the formula:
m=ρ·A·l
m= 2700 kg/m3×0.01 m2×1 m
m= 27 kg
Step 2: Calculate the increase in temperature. The change in temperature
(∆T) is given by:
∆T=Tfinal −Tinitial
∆T= 200◦C−100◦C
∆T= 100◦C
Step 3: Calculate the heat added to the rod. The heat added (Q) can be
calculated using the formula:
Q=m·c·∆T
where cis the specific heat capacity of aluminum, given as 900 J/(kg·K).
Q= 27 kg ×900 J/(kg ·K) ×100◦C
Q= 2,430,000 J
Therefore, the amount of heat added to the aluminum rod during this process
is 2,430,000 J.
Question 33
Question
A cube of aluminum with sides of length 10 cm at a temperature of 100
°
C
is submerged in a tank of water at 20
°
C. If the cube loses heat to the water
until it reaches thermal equilibrium, calculate the final temperature of the cube.
Assume specific heat capacity of aluminum is 900 J/kg ·K and the density of
aluminum is 2700 kg/m3. Also, assume that no heat is lost to the surroundings.
27
Solution
Step 1: Calculate the initial temperature of the aluminum cube in Kelvin. We
know that T(in Kelvin) = T(in Celsius) + 273.15. So, the initial temperature of
the aluminum cube is 373.15 K.
Step 2: Calculate the mass of the aluminum cube. The density of aluminum
is 2700 kg/m3, and the volume of the cube is 0.1 m ×0.1 m ×0.1 m = 0.001 m3.
Thus, the mass of the aluminum cube is 2700 kg/m3×0.001 m3= 2.7 kg.
Step 3: Calculate the heat lost by the aluminum cube. The heat lost by the
aluminum cube can be calculated using the formula Q=mc∆T, where: - m
is the mass of the aluminum cube (2.7 kg), - cis the specific heat capacity of
aluminum (900 J/kg ·K), - ∆Tis the change in temperature (final temperature
- initial temperature).
Step 4: Calculate the heat gained by the water. Since no heat is lost to the
surroundings, the heat lost by the aluminum cube is equal to the heat gained
by the water. We can use the formula Q=mc∆Tagain, where now: - mis
the mass of water (which we assume to be much greater than the mass of the
aluminum cube), - cis the specific heat capacity of water (4186 J/kg ·K), - ∆T
is the change in temperature (final temperature - initial water temperature of
20
°
C).
Step 5: Set up and solve an equation to find the final temperature of the
aluminum cube. Since the heat lost by the aluminum cube is equal to the
heat gained by the water, we can set up an equation: mc∆Tcube =mc∆Twater
Substitute the known values and solve for the final temperature of the aluminum
cube.
Step 6: Convert the final temperature from Kelvin to Celsius to find the
final temperature of the aluminum cube.
Question 34
Question
A piece of silver with a mass of 250 g and an initial temperature of 100
°
C is
placed in a container of 500 g of water at an initial temperature of 20
°
C. The
final temperature of the silver-water system is 25
°
C. Assuming no heat is lost
to the surroundings, calculate the specific heat capacity of silver. (Specific heat
capacity of water is 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the silver using the formula: Q=
mc∆T, where Qis the heat, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature. The heat lost by the silver (which decreases
its temperature) is equal to the heat gained by the water (which increases its
temperature), so we have: Qsilver =Qwater
28
Step 2: Calculate the heat lost or gained by the silver using the formula
Q=mc∆T. For the silver: Qsilver = (250 g)(csilver)(25 −100)
°
C
Step 3: Calculate the heat gained or lost by the water using the formula
Q=mc∆T. For the water: Qwater = (500 g)(4.18 J/g
°
C)(25 −20)
°
C
Step 4: Set the two heat equations from Steps 2 and 3 equal to each other
and solve for the specific heat capacity of silver: (250 g)(csilver)(25 −100) =
(500 g)(4.18 J/g
°
C)(25 −20)
Step 5: Solve for csilver:csilver =(500 g)(4.18 J/g
°
C)(25−20)
(250 g)(25−100)
Question 35
Question
A copper bar of mass 500 g at a temperature of 100
°
C is dropped into 500 g of
water at 20
°
C contained in a calorimeter of negligible heat capacity. Assuming
no heat is lost to the surroundings, what will be the final equilibrium temper-
ature of the system? (Specific heat capacity of copper = 0.385 J/g
°
C, specific
heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the copper bar as it cools down to the final
equilibrium temperature. The heat lost is given by the formula:
Qlost =m×c×∆T
where: - m= 500 g is the mass of the copper bar, - c= 0.385 J/g
°
C is the
specific heat capacity of copper, - ∆T= 100 −Tfis the temperature change,
and - Tfis the final equilibrium temperature.
Step 2: Calculate the heat gained by the water as it warms up to the final
equilibrium temperature. The heat gained is given by the formula:
Qgained =m×c×∆T
where: - m= 500 g is the mass of the water, - c= 4.18 J/g
°
C is the specific heat
capacity of water, - ∆T=Tf−20 is the temperature change.
Step 3: Since no heat is lost to the surroundings, the heat lost by the copper
bar must be equal to the heat gained by the water.
Qlost =Qgained
Step 4: Substitute the expressions for Qlost and Qgained into the equation
from Step 3 and solve for Tf.
m×ccopper ×(100 −Tf) = m×cwater ×(Tf−20)
500 ×0.385 ×(100 −Tf) = 500 ×4.18 ×(Tf−20)
29
Question 2
Question
A copper rod of length 1 m and initial temperature of 100
°
C is placed in contact
with a large heat reservoir at 0
°
C. The rod is allowed to come to thermal equilib-
rium with the reservoir. If the rod has a thermal conductivity of 385 W/(m
·
K)
and a cross-sectional area of 1 ×10−4m2, determine the rate at which heat is
transferred through the rod during this process.
Solution
Step 1: Calculate the temperature difference across the rod. Given that the
initial temperature of the rod is 100
°
C and it is in contact with a reservoir at
0
°
C, the temperature difference across the rod is:
∆T= 100C−0C= 100C
Step 2: Calculate the rate of heat transfer through the rod using Fourier’s
Law: The rate of heat transfer Qthrough the rod is given by:
Q=k·A·∆T
L
where: - kis the thermal conductivity of copper (385 W/(m
·
K)) - Ais the
cross-sectional area of the rod (1 ×10−4m2)-∆Tis the temperature difference
across the rod (100C) - Lis the length of the rod (1 m)
Plugging in the values:
Q= 385 W/(m
·
K) ×1×10−4m2×100C
1 m
Q= 38.5 W
Therefore, the rate at which heat is transferred through the rod during this
process is 38.5 W.
Question 3
Question
A copper rod of length 2 m has a cross-sectional area of 10 cm2. If one end of
the rod is at 100◦C and the other end is at 50◦C, calculate the rate of heat
conducted along the rod.
Given: Thermal conductivity of copper: 400 W/mK, Density of copper:
8900 kg/m3, Specific heat capacity of copper: 384 J/kgK.
2
Solution
Step 1: Calculate the heat conducted through the rod using Fourier’s law of
heat conduction:
Q=k·A·∆T
L
where: Q= rate of heat transfer (in watts), k= thermal conductivity of the
material (in W/mK), A= cross-sectional area of the material (in m2), ∆T=
change in temperature (in Kelvin), L= length of the material (in meters).
Given: k= 400 W/mK, A= 10 ×10−4m2, ∆T= 100◦C−50◦C = 50 K,
L= 2 m.
Substitute the values into the formula:
Q=400 ·10 ×10−4·50
2= 100 W
Therefore, the rate of heat conducted along the rod is 100 watts.
Question 4
Question
A copper block has a mass of 500 grams and is initially at a temperature of
100
°
C. It is placed in a bucket containing 2 kg of water at 20
°
C. Assuming no
heat is lost to the surroundings, calculate the final equilibrium temperature of
the system when the block and water reach thermal equilibrium. (Specific heat
capacity of copper = 0.386 J/g
°
C, specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the copper block Let Tbe the final
equilibrium temperature of the system. The heat lost by the copper block is
equal to the heat gained by the water:
(mcopper)(ccopper)(T−100) = (mwater)(cwater)(T−20)
Substitute the given values:
(500 g)(0.386 J/g
°
C)(T−100) = (2000 g)(4.18 J/g
°
C)(T−20)
Step 2: Solve for TExpand and simplify the equation:
193T−19300 = 8360T−167200
8360T−193T= 167200 −19300
8167T= 148900
T=148900
8167 ≈18.22C
Therefore, the final equilibrium temperature of the system when the block
and water reach thermal equilibrium is approximately 18.22C.
3
Question 5
Question
A copper bar of length 1.5 m and cross-sectional area 0.02 m2is heated from
20
°
C to 80
°
C. If the thermal conductivity of copper is 400 W/(m·K), calculate
the amount of heat transferred to the copper bar.
Solution
Step 1: Determine the initial and final temperatures of the copper bar in Kelvin.
Initial temperature (in K) = 20C+ 273.15 = 293.15K
Final temperature (in K) = 80C+ 273.15 = 353.15K
Step 2: Calculate the temperature difference (∆T) between the initial and
final temperatures.
∆T= 353.15K−293.15K
= 60K
Step 3: Calculate the heat transferred using the formula:
Q=k·A·∆T·t
where: Q= heat transferred (in Joules)
k= thermal conductivity of copper (400 W/(m·K))
A= cross-sectional area of the bar (0.02 m2)
∆T= temperature difference (60 K)
t= length of time of heat transfer (let’s assume 1 second for simplicity)
Step 4: Substitute the given values into the formula to find the heat trans-
ferred.
Q= 400 W/(m·K) ×0.02 m2×60 K ×1 s
= 480 J
Therefore, the amount of heat transferred to the copper bar is 480 Joules.
Question 6
Question
A metal bar of length 1.5 m and cross-sectional area 0.02 m2has a thermal
conductivity of 70 W/(m*K). One end of the bar is kept at 100◦C while the
other end is kept at 20◦C. Find the rate of heat transfer along the bar.
4
Solution
Step 1: Calculate the temperature difference along the bar. Given that the
metal bar has one end at 100◦C and the other end at 20◦C, the temperature
difference (∆T) is:
∆T= 100 −20 = 80 ◦C
Step 2: Calculate the rate of heat transfer using Fourier’s Law. Fourier’s
Law states that the rate of heat transfer ( ˙
Q) through a material is given by:
˙
Q=−kAdT
dx
where: ˙
Q= rate of heat transfer, k= thermal conductivity of the material, A
= cross-sectional area of the bar, dT
dx = temperature gradient along the bar.
Step 3: Calculate the temperature gradient using the temperature difference
and the length of the bar. The temperature gradient (dT
dx ) is given by:
dT
dx =∆T
L
where Lis the length of the bar. Substitute the values:
dT
dx =80 ◦C
1.5 m =80
1.5
◦C/m = 53.3◦C/m
Step 4: Substitute the values into Fourier’s Law to find the rate of heat
transfer. ˙
Q=−70 ×0.02 ×53.3
˙
Q=−70 ×0.02 ×53.3 = −74.6667 W
Therefore, the rate of heat transfer along the bar is 74.6667 W.
Question 7
Question
A block of metal initially at a temperature of 500 K is heated until its tem-
perature reaches 800 K. If the specific heat capacity of the metal is 0.5 J/g◦C,
calculate the amount of heat absorbed by a 250 g block of the metal during this
process.
Solution
Step 1: Calculate the change in temperature. Given: Initial temperature, T1=
500 K Final temperature, T2= 800 K Change in temperature, ∆T=T2−T1
∆T= 800 K −500 K = 300 K
5
Step 2: Calculate the amount of heat absorbed using the formula:
Q=mc∆T
where: Q= heat absorbed m= mass of the block = 250 g = 0.25 kg c=
specific heat capacity of the metal = 0.5 J/g◦C= 500 J/kg◦C
Substitute the values into the formula:
Q= 0.25 kg ×500 J/kg◦C×300 K
Step 3: Calculate the value of Q.
Q= 0.25 ×500 ×300 = 37,500 J
Therefore, the amount of heat absorbed by the block of metal is 37,500 J.
Question 8
Question
A block of copper with a mass of 0.5 kg and a specific heat capacity of 0.385
J/(g·
°
C) is initially at a temperature of 100
°
C. It is placed in a container with
2 kg of water at an initial temperature of 20
°
C. If the final temperature of the
system is 25
°
C, what is the specific heat capacity of water?
Solution
Step 1: Find the heat absorbed by the copper block using the formula:
Q=mc∆T
where: m= 0.5 kg (mass of copper), c= 0.385 J/(g ·
°
C) (specific heat capacity
of copper), ∆T= 25 −100 = −75
°
C (change in temperature).
Plugging in the values, we get:
Qcopper = (0.5 kg)(0.385 J/(g ·
°
C))(−75
°
C)
Qcopper =−14.375 J
Step 2: Find the heat released by the water when it cools down using the
same formula:
Q=mc∆T
where: m= 2 kg (mass of water), cis the specific heat capacity of water (to be
determined), ∆T= 25 −20 = 5
°
C (change in temperature).
Plugging in the values, we have:
−Qwater = (2 kg)(c)(5
°
C)
6
−Qwater = 10cJ
Step 3: Since the total heat gained by the water is equal to the total heat
lost by the copper block (by conservation of energy), we have:
Qcopper =Qwater
−14.375 J = 10cJ
c=−1.4375 J/(g ·
°
C)
Therefore, the specific heat capacity of water is 1.4375 J/(g·
°
C).
Question 9
Question
A block of copper initially at 100
°
C is dropped into a vessel containing 1 kg of
water at 20
°
C. The specific heat capacity of copper is 0.385 J/g
°
C and that of
water is 4.18 J/g
°
C. The block of copper comes to thermal equilibrium with the
water. Calculate the final temperature of the system.
Solution
Step 1: Find the heat lost by the copper block.
The heat lost by the copper block can be calculated using the formula:
Q=mc∆T, where: - mis the mass of the copper block, - cis the specific heat
capacity of copper, and - ∆Tis the change in temperature.
Given: - Initial temperature of copper block (Tcopper) = 100
°
C, - Final
temperature of system = Tf, - Specific heat capacity of copper (ccopper ) =
0.385 J/g
°
C, - Mass of copper block = mcopper, - Change in temperature (∆Tcopper)
=Tcopper −Tf.
The heat lost by the copper block is equal to the heat gained by the water,
so we have: mccopper ∆Tcopper =mcwater ∆Twater
Step 2: Find the heat gained by the water.
The heat gained by the water can be calculated using the same formula:
Q=mc∆T, where: - mis the mass of the water, - cis the specific heat capacity
of water, and - ∆Tis the change in temperature.
Given: - Initial temperature of water (Twater) = 20
°
C, - Mass of water = 1
kg = 1000 g, - Specific heat capacity of water (cwater) = 4.18 J/g
°
C, - Change
in temperature (∆Twater ) = Tf−Twater .
Step 3: Set up and solve the equations to find the final temperature.
From Step 1: mcopperccopper (Tcopper −Tf) = mwater cwater (Tf−Twater )
Substitute the given values and solve for Tfto find the final temperature of
the system.
7
Question 10
Question
A copper rod of length 2 m and diameter 1 cm conducts heat at a rate of 200 W
when its ends are at temperatures of 100◦C and 0◦C. The thermal conductivity
of copper is 400 W/m·K. How long does it take for the rod to reach a uniform
temperature of 50◦C?
Solution
Step 1: Find the cross-sectional area of the rod. The diameter of the rod is
1 cm, so the radius is 0.5 cm = 0.005 m. The cross-sectional area A of the
rod is given by the formula A=πr2, where r is the radius. A=π(0.005)2=
π×0.000025 = 7.85398 ×10−5m2.
Step 2: Find the rate of heat conduction per unit area. The rate of heat
conduction Q through a material is given by Fourier’s law: Q=kA∆T /L,
where k is the thermal conductivity, ∆Tis the temperature gradient, and L is
the length of the material. Given that Q = 200 W, k = 400 W/m·K, ∆T=
100 −0 = 100 ◦C = 100 K, and L = 2 m, we can rearrange the formula to find
Q
A=k∆T
L. Substitute the given values to find 200
7.85398×10−5= 400 ×100
2. This
simplifies to 200
7.85398×10−5= 200 ×104= 2 ×106W/m2.
Step 3: Find the time taken to reach a uniform temperature of 50◦C. The
rate of temperature change dT
dt at a point along the rod is given by Fourier’s law
in differential form: Q=−kA dT
dx , where x is the distance, in this case, along
the rod. At the midpoint of the rod, x = L/2 = 1 m. Solving for dT
dt , we get
dT
dt =−Q
kA =−2×106
400×7.85398×10−5=−50661430.7 K/s. To find the time taken
for the rod to reach a uniform temperature of 50◦C = 50 K, divide ∆Tby dT
dt .
The time taken is 50
50661430.7≈9.88 ×10−7s.
Therefore, it takes approximately 9.88 ×10−7seconds for the rod to reach
a uniform temperature of 50◦C.
Question 11
Question
A 150 g cube of ice at -10
°
C is placed in a 300 g aluminum calorimeter cup
of heat capacity 0.40 J/g
°
C containing 500 g of water at 20
°
C. Assume that
no heat is lost to the surroundings. What will be the final temperature of the
system?
Given: Specific heat capacity of ice: 2.09 J/g
°
C Specific heat capacity of
water: 4.18 J/g
°
C Heat of fusion of ice: 334 J/g
8
Solution
Step 1: Calculate the heat lost by the ice to reach 0
°
C. The heat lost by the ice
is given by the formula:
Qice =mice ·cice ·(0 −(−10))
Substitute the values:
Qice = 150 g ·2.09 J/g
°
C·10
°
C
Calculating:
Qice = 3135 J
Step 2: Calculate the heat gained by the ice at 0
°
C to melt into water. The
heat gained by the ice to melt is given by the formula:
Qmelt =mice ·Heat of fusion of ice
Substitute the given values:
Qmelt = 150 g ·334 J/g
Calculating:
Qmelt = 50100 J
Step 3: Calculate the final temperature after the ice has melted and reached
thermal equilibrium with the water. Let the final temperature be Tf.
Using the conservation of energy equation:
Qice +Qmelt = (mice ·cice +mcalorimeter ·ccalorimeter +mwater ·cwater)·(Tf−20)
Substitute the known values:
3135 + 50100 = (150 ·2.09 + 300 ·0.40 + 500 ·4.18) ·(Tf−20)
Solve for Tf:
53235 = (313.5 + 120 + 2090) ·(Tf−20)
53235 = 3523.5·(Tf−20)
Tf−20 = 53235
3523.5
Tf−20 ≈15
°
C
Therefore, the final temperature of the system is approximately 15
°
C.
9
Question 12
Question
A copper block with a mass of 0.5 kg is heated from an initial temperature of
25
°
C to a final temperature of 75
°
C. If the specific heat capacity of copper is
390 J/kg◦C, calculate the amount of heat energy absorbed by the block during
this process.
Solution
Step 1: Determine the change in temperature. Given: Initial temperature (Ti)
= 25
°
C, Final temperature (Tf) = 75
°
C.
The change in temperature (∆T) is calculated as:
∆T=Tf−Ti= 75◦C−25◦C = 50◦C
Step 2: Calculate the amount of heat energy absorbed. The amount of heat
energy absorbed (Q) is given by the formula:
Q=mc∆T
where: m= mass of the block = 0.5 kg, c= specific heat capacity of copper =
390 J/kg◦C.
Substitute the values into the formula:
Q= (0.5 kg)(390 J/kg◦C)(50◦C)
Q= 9750 J
Therefore, the amount of heat energy absorbed by the block during this
process is 9750 Joules.
Question 13
Question
A copper sphere with a radius of 5 cm is heated from an initial temperature
of 20
°
C to a final temperature of 80
°
C. If the specific heat capacity of copper
is 0.385 J/g
°
C and its density is 8.96 g/cm3, calculate the heat supplied to the
sphere in this process. Assume the sphere is perfectly insulated and no heat is
lost to the surroundings.
Solution
Step 1: Calculate the mass of the copper sphere using its density and volume.
Volume of sphere = 4
3πr3
10
Volume of sphere = 4
3π(5 cm)3
Volume of sphere = 4
3π×125 cm3
Volume of sphere = 500
3πcm3≈523.6 cm3
Mass = Volume ×Density
Mass = 523.6 cm3×8.96 g/cm3
Mass ≈4684.3 g = 4.6843 kg
Step 2: Calculate the change in temperature of the copper sphere.
∆T= 80C−20C= 60C
Step 3: Determine the heat supplied to the copper sphere using the formula:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
Q= 4.6843 ×0.385 ×60
Q≈108.01 kJ
Therefore, the heat supplied to the copper sphere in this process is approx-
imately 108.01 kJ.
Question 14
Question
A copper cylinder of mass 2 kg contains 0.5 kg of water at a temperature of
100◦C. The copper cylinder is then heated to a temperature of 200◦C. If the
specific heat capacity of copper is 390 J/kg·K and the specific heat capacity of
water is 4186 J/kg·K, determine the final temperature of the water and copper
cylinder system.
Solution
Step 1: Calculate the heat absorbed or released by the water using the formula:
Q=mc∆T
where Q is the heat energy, m is the mass, c is the specific heat capacity, and
∆Tis the change in temperature.
Given: - mwater = 0.5 kg - cwater = 4186 J/kg·K - Initial temperature of
water, Tin = 100◦C - Final temperature of water, Tfin
11
The change in temperature (∆T) for water is:
∆Twater =Tfin −Tin
The heat absorbed by water is:
Qwater =mwatercwater∆Twater
Step 2: Calculate the heat absorbed or released by the copper cylinder using
the same formula:
Q=mc∆T
Given: - mcopper = 2 kg - ccopper = 390 J/kg·K - Initial temperature of
copper, Tin = 100◦C - Final temperature of copper, Tfin = 200◦C
The change in temperature (∆T) for copper is:
∆Tcopper =Tfin −Tin
The heat absorbed by copper is:
Qcopper =mcopperccopper∆Tcopper
Step 3: Since the total heat energy absorbed by the water must equal the
total heat energy absorbed by the copper cylinder, we can set up an equation:
Qwater =Qcopper
Step 4: Substitute the expressions for Qwater and Qcopper and solve for Tfin.
Question 15
Question
A metal bar of length 2 m and cross-sectional area 0.02 m2is initially at a
uniform temperature of 200◦C. The bar is then placed in an environment at
20◦C, causing it to cool. If the thermal conductivity of the metal is 50 W/(m·K)
and the heat transfer coefficient between the bar and surrounding air is 25
W/(m2·K), determine the time it takes for the bar to cool to 100◦C. Assume
that the surroundings are at a constant temperature of 20◦C.
Solution
Step 1: Calculate the heat transfer rate from the bar to the surroundings using
Newton’s Law of Cooling: The heat transfer rate (Q) is given by:
Q=hA(T−Tsurroundings)
where h= heat transfer coefficient = 25 W/(m2·K), A= cross-sectional area =
0.02 m2,T= temperature of the bar at any time t,Tsurroundings = temperature
of the surroundings = 20◦C.
12
Step 2: Calculate the thermal resistance of the bar: The thermal resistance
(R) is given by:
R=L
kA
where L= length of the bar = 2 m, k= thermal conductivity = 50 W/(m·K),
A= cross-sectional area = 0.02 m2.
Step 3: Use the concept of thermal resistance to express the rate of tem-
perature change of the metal bar: The rate of temperature change is given
by: dT
dt =−Q
ρcV
where ρ= density of the material, c= specific heat capacity of the material, V
= volume of the bar.
Step 4: Substitute the expressions for Qand dT
dt into the rate of temperature
change equation and integrate to find the time tit takes for the bar to cool to
100◦C:
ZTf
Ti
dT
T−Tsurroundings
=−2
50 ·0.02 ·ρcV Zt
0
dt
where Ti= initial temperature of the bar = 200◦C, Tf= final temperature of
the bar = 100◦C.
Question 16
Question
A 0.5 kg aluminum pot containing 1 kg of water at 20
°
C is placed on a stove.
The stove supplies heat at a rate of 500 W. Assuming no heat is lost to the
surroundings, calculate how long it will take for the water to boil (100
°
C). The
specific heat capacity of aluminum is 900 J/kg◦C and the specific heat capacity
of water is 4186 J/kg◦C.
Solution
Step 1: Calculate the heat required to raise the temperature of water from 20
°
C
to 100
°
C. The heat Qrequired to raise the temperature of a material can be
calculated using the formula:
Q=mc∆T
where: - mis the mass of the material, - cis the specific heat capacity of the
material, - ∆Tis the change in temperature.
For the water:
Qwater = (1 kg)(4186 J/kg◦C)(80 ◦C)
Qwater = 334880 J
13
Step 2: Calculate the heat required to raise the temperature of the aluminum
pot from room temperature to the boiling point of water. For the aluminum
pot:
Qaluminum = (0.5 kg)(900 J/kg◦C)(80 ◦C)
Qaluminum = 36000 J
Step 3: Calculate the total heat provided by the stove to boil the water. The
total heat provided is the sum of the heat required for both the water and the
aluminum pot.
Qtotal =Qwater +Qaluminum
Qtotal = 334880 J + 36000 J = 370880 J
Step 4: Calculate the time taken for the water to boil. The rate of heat
supplied by the stove is given as 500 W (J/s). We can use the formula:
Power = Energy
Time
500 W = 370880 J
t
Solving for time t:
t=370880 J
500 W
t= 741.76 s
Therefore, it will take approximately 741.76 seconds for the water to boil.
Question 17
Question
A 2 kg block of ice at −20◦Cis placed in an insulated container holding 10 kg
of water at 25◦C. Assuming no heat is exchanged with the surroundings, what
will be the final temperature of the system? (Specific heat capacity of water =
4186 J/(kg ·K) and specific heat capacity of ice = 2100 J/(kg ·K)).
Solution
Step 1: First, we need to calculate the heat lost by the ice and the heat gained
by the water until they reach thermal equilibrium. The heat lost by the ice can
be calculated as:
Qice =mice ·cice ·(Tf−Tice)
where: - mice = 2 kg (mass of ice), - cice = 2100
Question
A copper rod with a length of 1.5 m and a diameter of 2 cm is initially at a
temperature of 200
°
C. It is placed in a room where the temperature is 25
°
C. If
14
the thermal conductivity of copper is 390 W/mK, the specific heat capacity is
390 J/kgK, and the density is 8900 kg/m3, calculate the time it takes for the
rod to reach thermal equilibrium with the room. Assume one end of the rod is
perfectly insulated.
Solution
Step 1: Calculate the surface area of the rod. The surface area of a rod can be
calculated using the formula A=π×r×h, where ris the radius and his the
length of the rod. Given that the diameter of the rod is 2 cm, the radius ris 1
cm or 0.01 m. So, A=π×0.01 ×1.5 = 0.0471 m2.
Step 2: Calculate the volume of the rod. The volume of the rod can be
calculated using the formula V=π×r2×h. Given that the diameter of the
rod is 2 cm, the radius ris 1 cm or 0.01 m. So, V=π×(0.01)2×1.5 = 4.712
x 10−4m3.
Step 3: Calculate the mass of the rod. The mass of the rod can be calculated
using the formula m= density ×volume. Given that the density of copper is
8900 kg/m3, the mass m= 8900 ×4.712 x 10−4= 4.2 kg.
Step 4: Calculate the total energy required to heat the rod from 25
°
C to
200
°
C. The total energy Qrequired can be calculated using the formula Q=
mc∆T, where mis the mass, cis the specific heat capacity, and ∆Tis the
temperature difference. The temperature difference ∆T= 200 −25 = 175
°
C.
So, Q= 4.2×390 ×175 = 296550 J.
Step 5: Calculate the rate of heat transfer through the rod. The rate of heat
transfer Pcan be calculated using Fourier’s law: P=kA∆T/L, where kis the
thermal conductivity, Ais the surface area, ∆Tis the temperature difference,
and Lis the length of the rod. One end of the rod is insulated, so only one side is
in contact with the room. Therefore, L= 1.5 m. So, P= 390×0.0471×175/1.5
= 21.64 W.
Step 6: Calculate the time taken for the rod to reach thermal equilibrium.
The time ttaken can be calculated using the formula Q=P t, where Qis the
total energy required, and Pis the rate of heat transfer. So, t= 296550/21.64
= 13706 seconds or 3.81 hours. Therefore, it takes approximately 3.81 hours for
the rod to reach thermal equilibrium with the room.
Question 19
Question
A piece of aluminum initially at 200
°
C is placed in a lake with a temperature of
10
°
C. If the specific heat capacity of aluminum is 0.897 J/g
°
C, and the mass of
the aluminum is 500 g, calculate the amount of heat that must be transferred
to the aluminum to bring it to thermal equilibrium with the lake.
15
Solution
Step 1: Calculate the change in temperature of the aluminum. Let Tfbe the
final temperature of the aluminum. The change in temperature ∆Tis given by:
∆T=Tf−Ti
where Ti= 200Cis the initial temperature of the aluminum. Therefore, ∆T=
Tf−200.
Step 2: Calculate the heat absorbed or lost by the aluminum using the
formula:
q=mc∆T
where qis the heat, mis the mass of the aluminum, cis the specific heat capacity
of aluminum, and ∆Tis the change in temperature.
Step 3: Set up the equation using the given values:
q= (500 g)(0.897 J/g
°
C)(Tf−200)
Step 4: At thermal equilibrium, the total heat loss by the aluminum equals
the total heat gained by the lake. This means:
q=mc∆T=−qlake
where qlake is the heat lost by the lake (negative since the lake gains heat).
Step 5: Since the lake is at 10
°
C, the heat lost by the lake is given by:
−qlake = 500 g ×(specific heat capacity of water) ×(Tf−10)
Step 6: Equate qand −qlake:
500 ×0.897 ×(Tf−200) = −500 ×(specific heat capacity of water) ×(Tf−10)
Step 7: Solve for Tfto find the final temperature of the aluminum when it
reaches thermal equilibrium with the lake. After finding Tf, substitute it back
into the equation for qto calculate the amount of heat transferred.
Question 20
Question
A metal bar of length 2 m and cross-sectional area 0.01 m2is heated at one end
and the other end is kept at a constant temperature. The thermal conductivity
of the metal is 200 W/(m K). If the temperature gradient along the length of
the bar is 100 K/m, determine the rate at which energy is conducted along the
bar.
16
Solution
Let’s denote the rate at which energy is conducted along the bar by Q. We can
use Fourier’s law of heat conduction which states:
Q=−kA∆T
∆x
where: Q= rate of heat conduction, k= thermal conductivity of the material, A
= cross-sectional area, ∆T= change in temperature, ∆x= change in position.
Step 1: Find the change in position ∆x.Given that the length of the
bar is 2 m and the temperature gradient is 100 K/m, we can find the change in
position as:
∆x= 2 m
Step 2: Calculate the rate of heat conduction. Plugging in the values
into Fourier’s law of heat conduction, we get:
Q=−200 ×0.01 ×100 = −200 W
Therefore, the rate at which energy is conducted along the bar is 200 W.
Question 21
Question
A copper block of mass 0.5 kg at an initial temperature of 200◦C is placed in
a calorimeter containing 0.2 kg of water at 20◦C. If the final temperature of
the system is 50◦C, calculate the specific heat capacity of copper. Assume the
specific heat capacity of water is 4200 J/kg·K.
Solution
Step 1: Calculate the heat gained by the water. The heat gained by the water
can be calculated using the formula:
Qwater =m·c·∆T
where: - mis the mass of the water, which is 0.2 kg, - cis the specific heat
capacity of water, which is 4200 J/kg·K, - ∆Tis the change in temperature of
the water, which is 50◦C - 20◦C = 30◦C.
Substitute the given values into the formula:
Qwater = 0.2 kg ×4200 J/kg·K×30 ◦C
Step 2: Calculate the heat lost by the copper block. The heat lost by the
copper block can also be calculated using the formula:
Qcopper =m·c·∆T
17
where: - mis the mass of the copper block, which is 0.5 kg, - cis the specific
heat capacity of copper (to be calculated), - ∆Tis the change in temperature of
the copper block, which is 50◦C - 200◦C = -150◦C (negative because the block
is losing heat).
Step 3: The total heat lost by the copper block is equal to the total heat
gained by the water.
Qcopper =Qwater
mc·cc·∆Tc=mw·cw·∆Tw
Solve for ccto find the specific heat capacity of copper:
cc=mw·cw·∆Tw
mc·∆Tc
Substitute the known values into the formula and solve for cc.
Question 22
Question
A 10 kg block of copper initially at a temperature of 100◦C is dropped into a
container of 20 kg of water at an initial temperature of 20◦C. Assuming no heat
loss to the surroundings, calculate the final equilibrium temperature of the block
and water system. (Specific heat capacity of copper = 390 J/kg◦C, specific heat
capacity of water = 4186 J/kg◦C)
Solution
Step 1: Calculate the heat lost by the copper block as it cools down to the final
temperature of the system. The heat lost by an object can be calculated using
the formula:
Q=mc∆T
where: Q= heat lost, m= mass of the object, c= specific heat capacity of the
object, ∆T= change in temperature.
For the copper block: m= 10 kg, c= 390 J/kg◦C, Initial temperature =
100◦C, Final temperature = T(since the final equilibrium temperature is not
known yet).
We can then calculate the heat lost by the copper block:
Qcopper = 10 ×390 ×(100 −T)
Step 2: Calculate the heat gained by the water as it heats up to the final
temperature of the system. The heat gained by an object can be calculated
using the same formula as in Step 1.
For the water: m= 20 kg, c= 4186 J/kg◦C, Initial temperature = 20◦C,
Final temperature = T.
18
We can calculate the heat gained by the water:
Qwater = 20 ×4186 ×(T−20)
Step 3: Since the system reaches thermal equilibrium, the heat lost by the
copper block is equal to the heat gained by the water.
Qcopper =Qwater
10 ×390 ×(100 −T) = 20 ×4186 ×(T−20)
Now, we can solve this equation to find the final equilibrium temperature T
of the system.
Question 23
Question
A metal bar of length 2.0 m and cross-sectional area 5.0 cm2is initially at a
temperature of 100◦C. If 5.0 kJ of heat are added to the bar, what is the final
temperature of the bar? Assume the specific heat capacity of the metal is 0.450
kJ/kg◦C and the density is 8.0 g/cm3.
Solution
Step 1: Calculate the mass of the metal bar using its length, cross-sectional
area, and density. The volume of the bar is given by:
V= length ×cross-sectional area = 2.0×5.0×10−4= 0.001 m3
The mass can be calculated from the density:
mass = density ×volume = 8.0×103×0.001 = 8.0 kg
Step 2: Calculate the change in temperature of the metal bar. The heat
added to the bar is given as 5.0 kJ, and the specific heat capacity is 0.450
kJ/kg◦C. So, the change in temperature can be calculated using the formula:
Q=mc∆T
where: Q= heat added = 5000 J m= mass = 8.0 kg c= specific heat capacity
= 450 J/kg◦C
Solving for ∆T:
5000 = 8.0×450 ×∆T
∆T=5000
8.0×450 = 1.389 ◦C
Step 3: Calculate the final temperature of the metal bar. The final temper-
ature can be found by adding the change in temperature to the initial temper-
ature:
Final temperature = 100 + 1.389 = 101.39 ◦C
Therefore, the final temperature of the metal bar after adding 5.0 kJ of heat
is 101.39◦C.
19
Question 24
Question
A copper rod of length 1.5 m and diameter 1 cm is initially at a temperature of
25
°
C. If one end of the rod is immersed in steam at 100
°
C and the other end is
surrounded by ice at 0
°
C, what is the rate at which energy is transferred along
the rod? Given that the thermal conductivity of copper is 390 W/mK.
Solution
Step 1: First, we calculate the cross-sectional area of the rod. The radius of
the rod, r=1 cm
2= 0.005 m. Therefore, the cross-sectional area of the rod,
A=πr2=π(0.005)2m2.
Step 2: Next, we calculate the temperature difference across the rod. The
temperature difference across the rod, ∆T= (100 −25) −(25 −0) = 75
°
C.
Step 3: Now, we can calculate the rate of energy transferred along the rod
using Fourier’s Law of Heat Conduction:
dQ
dt =−kAdT
dx
where, dQ/dt = rate of heat transfer, k= thermal conductivity of copper, A=
cross-sectional area of the rod, dT/dx = temperature gradient along the rod.
Step 4: The temperature gradient along the rod, dT /dx =∆T
L=75
1.5
°
C/m.
Step 5: Substituting the given values into Fourier’s Law, we get:
dQ
dt =−390 ×π×(0.005)2×75
1.5
Step 6: Simplifying,
dQ
dt =−390 ×π×0.000025 ×50
Step 7: Finally, calculating the rate of energy transferred gives:
dQ
dt =−π×0.4875
Therefore, the rate at which energy is transferred along the rod is approxi-
mately −1.53 W/m.
Question 25
Question
A metal rod of length 1 m and diameter 2 cm is initially at a temperature of
100◦C. It is then immersed in a large tank of water at 20◦C. If the rod loses heat
at a rate of 50 W, how long will it take for the rod to cool down to 50◦C? Assume
that the thermal conductivity of the metal is 100 W/mK and the specific heat
capacity of the metal is 500 J/kgK.
20
Solution
Step 1: Calculate the surface area of the rod. The surface area of a cylinder can
be calculated using the formula:
A= 2πrh +πr2
where ris the radius of the cylinder and his its height. In this case, h= 1 m
and r= 1 cm = 0.01 m.
So, the surface area of the rod is:
A= 2π(0.01)(1) + π(0.01)2= 0.04 m2
Step 2: Determine the rate of temperature change. The rate at which the
rod loses heat is given as 50 W. We can calculate this rate using the formula:
Q=kA∆T
∆x
where Qis the heat flow rate, kis the thermal conductivity of the metal, Ais
the surface area of the rod, ∆Tis the temperature difference, and ∆xis the
length of the rod.
Substitute the given values into the equation:
50 = (100)(0.04)(100 −20)
1
Step 3: Calculate the time taken to cool to 50◦C. First, we need to calculate
the mass of the rod using the density formula:
Density = m
Volume
m= Density ×Volume
The volume of the rod is given by:
V=πr2h=π(0.01)2(1) = 0.0001πm3
So, the mass of the rod is:
m= 8000 ×0.0001π= 0.25πkg
The specific heat formula is:
Q=mc∆T
50 ×t= 0.25π×500 ×(100 −50)
50t= 6250π
t≈39.8 s
Therefore, it will take approximately 39.8 seconds for the rod to cool down
to 50◦C.
21
Question 26
Question
A copper cubical block with sides of length 0.1 m is initially at a temperature
of 100
°
C. It is dropped into a large container of cold water at 10
°
C. If the
specific heat capacity of copper is 390 J/kg ·
°
C and the density of copper is
8,960 kg/m3, calculate the total heat transfer when the block reaches thermal
equilibrium with the water. Assume no heat is lost to the surroundings.
Solution
Step 1: Calculate the mass of the copper block. Given that the density of
copper is 8,960 kg/m3and the volume of the copper block is (0.1 m)3, we can
calculate the mass using the formula m=ρV .m= 8,960 kg/m3×(0.1 m)3
m= 8,960 kg/m3×0.001 m3m= 8.96 kg
Step 2: Calculate the heat lost by the copper block. The heat lost by the
copper block can be calculated using the formula Q=mc∆T, where mis the
mass, cis the specific heat capacity, and ∆Tis the change in temperature.
∆T= 100C−10C= 90C Q = 8.96 kg ×390 J/kg ·
°
C×90
°
CQ= 316,872 J
Step 3: Calculate the heat gained by the water. The heat gained by the
water can be calculated using the same formula Q=mc∆T, where the mass
of the water and the specific heat capacity of water are used. Let’s assume the
mass of the water is equal to the mass of the copper block. Q= 8.96 kg ×
4180 J/kg ·
°
C×90
°
CQ= 337,843.2 J
Step 4: Total heat transfer. Since the heat lost by the copper block is equal
to the heat gained by the water, the total heat transfer is the sum of these two
amounts. Total heat transfer = 316,872 J + 337,843.2 J = 654,715.2 J
Therefore, the total heat transfer when the copper block reaches thermal
equilibrium with the water is 654,715.2 J.
Question 27
Question
A copper vessel of mass 500 g contains 2 kg of water at 30
°
C. How much ice at
-10
°
C must be added to it so that the final temperature of the mixture becomes
10
°
C?
Solution
Step 1: Calculate the heat lost by the water to cool down from 30
°
C to 10
°
C.
Let mwbe the mass of the water, Tibe the initial temperature of the water,
Tfbe the final temperature of the water, and cwbe the specific heat capacity
of water. The specific heat capacity of water is 4200 J/kg
°
C. The formula for
22
calculating heat transfer is given by:
Q=mc∆T
where: - Qis the heat transfer (in Joules), - mis the mass (in kg), - cis the
specific heat capacity (in J/kg
°
C), and - ∆Tis the change in temperature (in
°
C).
The heat lost by the water is equal to the heat gained by the ice. Therefore,
we can write:
mwcw(Tf−Ti) = miLf+mici(Tf−Ti)
where: - miis the mass of the ice added, - Lfis the latent heat of fusion of ice
(336,000 J/kg), and - ciis the specific heat capacity of ice (2100 J/kg
°
C).
Step 2: Substituting the known values into the equation, we get:
mw·4200 ·(10 −30) = mi·336000 + mi·2100 ·(10 −(−10))
Step 3: Simplify the equation to solve for mi:
2·4200 ·20 = mi·336000 + mi·2100 ·20
Step 4: Solve for mito find the mass of ice needed to be added to the
mixture.
Question 28
Question
A steel rod of length 2 m and diameter 1 cm is heated from 20
°
C to 100
°
C. If the
coefficient of linear expansion of steel is 12 ×10−6per degree Celsius, calculate
the change in length of the rod.
Solution
Step 1: First, we calculate the initial volume of the steel rod. The initial volume
Vican be calculated using the formula for the volume of a cylinder:
Vi=πd
22
·L,
where dis the diameter of the rod and Lis the length of the rod. Substituting
d= 1 cm, L= 2 m, we get:
Vi=π1
22
·2 = π
2m3.
Step 2: Next, we calculate the final volume of the steel rod. We know that
volume is conserved, so the final volume Vfis equal to the initial volume Vi.
23
The final volume Vfcan be calculated using the formula for the volume of a
cylinder with the new length Lfand the same diameter:
Vf=πd
22
·Lf.
Since Vi=Vf, we can write:
π
2=π1
22
·Lf,
Lf= 4 m.
Step 3: Finally, we calculate the change in length of the rod. The change in
length ∆Lis given by:
∆L=Lf−Li,
∆L= 4 m −2 m,
∆L= 2 m.
Therefore, the change in length of the steel rod when heated from 20
°
C to
100
°
C is 2 meters.
Question 29
Question
A certain object is initially at a temperature of 300 K. If heat is added to
the object at a rate of 50 J/s for 10 minutes, what is the final temperature of
the object? Assume no heat is lost to the surroundings and the specific heat
capacity of the object is 500 J/kg-K.
Solution
Step 1: First, we need to convert the time from minutes to seconds. There are
60 seconds in a minute, so 10 minutes is equal to 600 seconds.
Step 2: Next, we calculate the total heat added to the object using the
formula for heat: Q= rate of heat transfer ×time. Substituting in the known
values, we get:
Q= 50 J/s ×600 s = 30,000 J
Step 3: The change in temperature of the object can be calculated using the
formula ∆T=Q
mc , where mis the mass of the object and cis the specific heat
capacity. Since the mass is not given and does not affect the final temperature,
we can proceed with the calculation:
∆T=30,000 J
m×500 J/kg-K
24
Step 4: The final temperature can be found by adding the change in tem-
perature to the initial temperature. Thus, the final temperature is:
Tfinal = 300 K + ∆T
Step 5: Substituting the value of ∆Tcalculated in step 3 into the equation
from step 4, we get:
Tfinal = 300 K + 30,000 J
m×500 J/kg-K
Question 30
Question
A copper rod of length 1 m and cross-sectional area 0.01 m2is initially at
100◦C. The rod is placed in an ice-water mixture at 0◦C and allowed to reach
thermal equilibrium with the surroundings. Given that copper has a specific
heat capacity of 390 J/(kg·K) and a density of 8930 kg/m3, calculate the amount
of heat transferred from the rod to the surroundings. Assume no heat is lost to
the surroundings other than through the ends of the rod, and that the specific
heat capacity and density of copper remain constant over the temperature range.
Solution
Step 1: Calculate the mass of the copper rod. The mass of the rod can be
calculated using the formula:
mass = density ×volume
Given that the density of copper is 8930 kg/m3and the volume of the rod is
length ×cross-sectional area, we have:
mass = 8930 kg/m3×1 m ×0.01 m2= 89.3 kg
Step 2: Calculate the initial thermal energy of the rod. The initial thermal
energy Qiof the rod can be calculated using the formula:
Qi= mass ×specific heat capacity ×∆T
where ∆Tis the change in temperature. The rod is initially at 100◦C, so
∆T= 100 −0 = 100 K. Substitute the values to get:
Qi= 89.3 kg ×390 J/(kg ·K) ×100 K = 349470 J
Step 3: Calculate the final thermal energy of the rod. When the rod reaches
thermal equilibrium at 0◦C, the final thermal energy Qfof the rod is:
Qf= mass ×specific heat capacity ×∆T
25
where ∆T= 0 −100 = −100 K. Substitute the values to get:
Qf= 89.3 kg ×390 J/(kg ·K) ×(−100) K = −349470 J
Step 4: Calculate the amount of heat transferred. The amount of heat
transferred from the rod to the surroundings can be calculated as the decrease
in thermal energy:
Q=Qf−Qi=−349470 J −349470 J = −698940 J
Therefore, the amount of heat transferred from the rod to the surroundings is
698940 J.
Question 31
Question
A copper cylinder of mass 0.2 kg is initially at 100
°
C. It is placed in a large heat
sink at 0
°
C. If the specific heat capacity of copper is 390 J/kg◦C, calculate the
total heat transfer when the cylinder reaches thermal equilibrium with the heat
sink.
Solution
Step 1: Calculate the change in temperature of the copper cylinder. The heat
transferred by the copper cylinder can be given by the formula:
Q=mc∆T,
where: Q= heat transferred, m= mass of copper cylinder = 0.2 kg, c= specific
heat capacity of copper = 390 J/kg◦C, ∆T= change in temperature.
Initially, the temperature of the copper cylinder is 100
°
C, and the final tem-
perature is 0
°
C. Therefore, the change in temperature is:
∆T= 0 −100 = −100 ◦C.
Step 2: Calculate the heat transferred. Substitute the values into the formula
for heat transferred:
Q= (0.2)(390)(−100).
Calculating this gives:
Q=−7800 J.
Step 3: Interpretation. The negative sign indicates that heat is being trans-
ferred out of the copper cylinder to the heat sink since the final temperature is
lower than the initial temperature. Therefore, the total heat transfer when the
cylinder reaches thermal equilibrium with the heat sink is 7800 J.
26
Question 32
Question
An aluminum rod of length 1 m and cross-sectional area 0.01 m2is initially at
a temperature of 100◦C. The rod is heated until its temperature reaches 200◦C.
The specific heat capacity of aluminum is 900 J/(kg·K) and its density is 2700
kg/m3. How much heat is added to the rod during this process?
Solution
Step 1: Calculate the mass of the aluminum rod. Given the length of the rod
(l) is 1 m, the cross-sectional area (A) is 0.01 m2, and the density of aluminum
(ρ) is 2700 kg/m3, we can calculate the mass (m) of the rod using the formula:
m=ρ·A·l
m= 2700 kg/m3×0.01 m2×1 m
m= 27 kg
Step 2: Calculate the increase in temperature. The change in temperature
(∆T) is given by:
∆T=Tfinal −Tinitial
∆T= 200◦C−100◦C
∆T= 100◦C
Step 3: Calculate the heat added to the rod. The heat added (Q) can be
calculated using the formula:
Q=m·c·∆T
where cis the specific heat capacity of aluminum, given as 900 J/(kg·K).
Q= 27 kg ×900 J/(kg ·K) ×100◦C
Q= 2,430,000 J
Therefore, the amount of heat added to the aluminum rod during this process
is 2,430,000 J.
Question 33
Question
A cube of aluminum with sides of length 10 cm at a temperature of 100
°
C
is submerged in a tank of water at 20
°
C. If the cube loses heat to the water
until it reaches thermal equilibrium, calculate the final temperature of the cube.
Assume specific heat capacity of aluminum is 900 J/kg ·K and the density of
aluminum is 2700 kg/m3. Also, assume that no heat is lost to the surroundings.
27
Solution
Step 1: Calculate the initial temperature of the aluminum cube in Kelvin. We
know that T(in Kelvin) = T(in Celsius) + 273.15. So, the initial temperature of
the aluminum cube is 373.15 K.
Step 2: Calculate the mass of the aluminum cube. The density of aluminum
is 2700 kg/m3, and the volume of the cube is 0.1 m ×0.1 m ×0.1 m = 0.001 m3.
Thus, the mass of the aluminum cube is 2700 kg/m3×0.001 m3= 2.7 kg.
Step 3: Calculate the heat lost by the aluminum cube. The heat lost by the
aluminum cube can be calculated using the formula Q=mc∆T, where: - m
is the mass of the aluminum cube (2.7 kg), - cis the specific heat capacity of
aluminum (900 J/kg ·K), - ∆Tis the change in temperature (final temperature
- initial temperature).
Step 4: Calculate the heat gained by the water. Since no heat is lost to the
surroundings, the heat lost by the aluminum cube is equal to the heat gained
by the water. We can use the formula Q=mc∆Tagain, where now: - mis
the mass of water (which we assume to be much greater than the mass of the
aluminum cube), - cis the specific heat capacity of water (4186 J/kg ·K), - ∆T
is the change in temperature (final temperature - initial water temperature of
20
°
C).
Step 5: Set up and solve an equation to find the final temperature of the
aluminum cube. Since the heat lost by the aluminum cube is equal to the
heat gained by the water, we can set up an equation: mc∆Tcube =mc∆Twater
Substitute the known values and solve for the final temperature of the aluminum
cube.
Step 6: Convert the final temperature from Kelvin to Celsius to find the
final temperature of the aluminum cube.
Question 34
Question
A piece of silver with a mass of 250 g and an initial temperature of 100
°
C is
placed in a container of 500 g of water at an initial temperature of 20
°
C. The
final temperature of the silver-water system is 25
°
C. Assuming no heat is lost
to the surroundings, calculate the specific heat capacity of silver. (Specific heat
capacity of water is 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat gained or lost by the silver using the formula: Q=
mc∆T, where Qis the heat, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature. The heat lost by the silver (which decreases
its temperature) is equal to the heat gained by the water (which increases its
temperature), so we have: Qsilver =Qwater
28
Step 2: Calculate the heat lost or gained by the silver using the formula
Q=mc∆T. For the silver: Qsilver = (250 g)(csilver)(25 −100)
°
C
Step 3: Calculate the heat gained or lost by the water using the formula
Q=mc∆T. For the water: Qwater = (500 g)(4.18 J/g
°
C)(25 −20)
°
C
Step 4: Set the two heat equations from Steps 2 and 3 equal to each other
and solve for the specific heat capacity of silver: (250 g)(csilver)(25 −100) =
(500 g)(4.18 J/g
°
C)(25 −20)
Step 5: Solve for csilver:csilver =(500 g)(4.18 J/g
°
C)(25−20)
(250 g)(25−100)
Question 35
Question
A copper bar of mass 500 g at a temperature of 100
°
C is dropped into 500 g of
water at 20
°
C contained in a calorimeter of negligible heat capacity. Assuming
no heat is lost to the surroundings, what will be the final equilibrium temper-
ature of the system? (Specific heat capacity of copper = 0.385 J/g
°
C, specific
heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the copper bar as it cools down to the final
equilibrium temperature. The heat lost is given by the formula:
Qlost =m×c×∆T
where: - m= 500 g is the mass of the copper bar, - c= 0.385 J/g
°
C is the
specific heat capacity of copper, - ∆T= 100 −Tfis the temperature change,
and - Tfis the final equilibrium temperature.
Step 2: Calculate the heat gained by the water as it warms up to the final
equilibrium temperature. The heat gained is given by the formula:
Qgained =m×c×∆T
where: - m= 500 g is the mass of the water, - c= 4.18 J/g
°
C is the specific heat
capacity of water, - ∆T=Tf−20 is the temperature change.
Step 3: Since no heat is lost to the surroundings, the heat lost by the copper
bar must be equal to the heat gained by the water.
Qlost =Qgained
Step 4: Substitute the expressions for Qlost and Qgained into the equation
from Step 3 and solve for Tf.
m×ccopper ×(100 −Tf) = m×cwater ×(Tf−20)
500 ×0.385 ×(100 −Tf) = 500 ×4.18 ×(Tf−20)
29
Step 5: Solve the equation for Tf.
192.5(100 −Tf) = 2090(Tf−20)
19250 −192.5Tf= 2090Tf−41800
22790 = 2282.5Tf
Tf=22790
2282.5≈10
°
C
Therefore, the final equilibrium temperature of the system will be approxi-
mately 10
°
C.
30
